A triangle has corners at (9 ,5 ), (2 ,3 ), and (7 ,6 ). What is the area of the triangle's circumscribed circle?

Oct 15, 2017

$\text{Area} = \frac{4505}{242} \pi$

Explanation:

The standard Cartesian form for the equation of a circle is:

${\left(x - h\right)}^{2} + {\left(y - k\right)}^{2} = {r}^{2} \text{ [1]}$

where $\left(x , y\right)$ is any point on the circle, $\left(h , k\right)$ is the center point, and $r$ is the radius.

We can use the points, $\left(9 , 5\right) , \left(2 , 3\right) , \left(7 , 6\right)$, and equation [1] to write 3 equations:

${\left(9 - h\right)}^{2} + {\left(5 - k\right)}^{2} = {r}^{2} \text{ [2]}$
${\left(2 - h\right)}^{2} + {\left(3 - k\right)}^{2} = {r}^{2} \text{ [3]}$
${\left(7 - h\right)}^{2} + {\left(6 - k\right)}^{2} = {r}^{2} \text{ [4]}$

Expand the squares:

$81 - 18 h + {h}^{2} + 25 - 10 k + {k}^{2} = {r}^{2} \text{ [2.1]}$
$4 - 4 h + {h}^{2} + 9 - 6 k + {k}^{2} = {r}^{2} \text{ [3.1]}$
$49 - 14 h + {h}^{2} + 36 - 12 k + {k}^{2} = {r}^{2} \text{ [4.1]}$

Subtract equation [3.1] from equation [2.1]:

$81 - 18 h + {h}^{2} + 25 - 10 k + {k}^{2} = {r}^{2}$
$\underline{- 4 + 4 h - {h}^{2} - 9 + 6 k - {k}^{2} = - {r}^{2}}$
$77 - 14 h - 0 {h}^{2} + 16 - 4 h + 0 {k}^{2} = 0$

Combine like terms:

$93 - 14 h - 4 k = 0 \text{ [5]}$

Subtract equation [3.1] from equation [4.1]:

$49 - 14 h + {h}^{2} + 36 - 12 k + {k}^{2} = {r}^{2}$
$\underline{- 4 + 4 h - {h}^{2} - 9 + 6 k - {k}^{2} = - {r}^{2}}$
$45 + 10 h + 0 {h}^{2} + 27 + 6 k + 0 {k}^{2} = 0$

Combine like terms:

$72 - 10 h - 6 k = 0 \text{ [6]}$

Multiply both sides of equation [5] by $- \frac{3}{2}$ and add to equation [6]:

72 - 10h -6k - 3/2(93 - 14h - 4k) = 0

Distribute the $- \frac{3}{2}$:

72 - 10h -6k - 279/2 + 21h + 6k = 0

Combine like terms:

$11 h - \frac{135}{2}$

$h = \frac{135}{22}$

Use equation [6] to find the value of k:

$72 - 10 \left(\frac{135}{22}\right) - 6 k = 0 \text{ [6]}$

$\frac{234}{22} - 6 k = 0$

$k = \frac{39}{22}$

Use equation [2] to find the value of ${r}^{2}$

${\left(9 - \frac{135}{22}\right)}^{2} + {\left(5 - \frac{39}{22}\right)}^{2} = {r}^{2}$

${r}^{2} = {\left(\frac{63}{22}\right)}^{2} + {\left(\frac{71}{22}\right)}^{2}$

${r}^{2} = \frac{9010}{484}$

${r}^{2} = \frac{4505}{242}$

Because the area of a circle is $\pi {r}^{2}$ we only need to multiply by $\pi$:

$\text{Area} = \frac{4505}{242} \pi$

Oct 15, 2017

Circum center (6.1364, 1.7727)
Area of circumcircle 58.5065

Explanation:

Slope of AB m1 = (3-5)/(2-9)=2/7.
Slope of perpendicular at mid point of AB = -1/m1 = -7/2
Midpoint of AB = (9+2)/2, (3+5)/2 = 11/2, 4
Eqn of perpendicular bisector of AB is
$y - 4 = - \left(\frac{7}{2}\right) \left(x - \left(\frac{11}{2}\right)\right)$
$14 x + 4 y = 93 \textcolor{w h i t e}{a a a} E q n \left(1\right)$

Slope of BC m2 = (6-3)/(7-2) = 3/5.
Slope of perpendicular at mid point of AB = -1/m1 = -5/3.
Midpoint of BC = (7+2)/2, (6+3)/2 = 9/2, 9/2
Eqn of perpendicular bisector of BC is
$y - \frac{9}{2} = \left(- \frac{5}{3}\right) \left(x - \left(\frac{9}{2}\right)\right)$
$10 x + 6 y = 72 \textcolor{w h i t e}{a a a} E q n \left(2\right)$

Solving Eqns (1), (2)
$x = \left(\frac{135}{22}\right) , y = \left(\frac{39}{22}\right)$
Circum center (6.1364, 1.7727)

r^2 = (9-(135/22))^2+(5-(39/22))^2 r^2=(63^2+71^2)/22^2#

Area of circumcircle $= \pi \cdot {r}^{2} = \frac{\cancel{22} \cdot \left({63}^{2} + {71}^{2}\right)}{7 \cdot \cancel{22} \cdot 22}$
$= 58.5065$